but since h must also be a multiplicative homomorphism, we need h(k) = h(k*1) = h(k)h(1), that is: h(1) must be the (unique!) multiplicative identity of h(Z). so if h(Z) = Z/30Z, for example, we have to have h(1) = 1.

this severely cuts down on the number of possible ring homomorphisms.

for example, we can't have h(1) = 2, since we must have h(1) = h(1)h(1), but if h(1) = 2, h(1)h(1) = 2*2 = 4.

it is clear that each additive subgroup can have at most one multiplicative identity. some trial-and-error shows that 16 is such an identity for <2>. so we need to verify that:

note that all we are doing is sending k (mod 30) to k (mod 15). so even though it looks like we are "jumbling" the order, all the additive and multiplicative properties of the ring Z/15Z are preserved:

i suspect that what is actually giving you trouble here, is that you aren't fully grasping how quotient groups work. the multiplicative structure of Z is built upon the additive structure of Z (it's the analogue of "powers" in a "generic" group). the multiplicative structure of Z/nZ is inherited from that of Z. in particular, the finer structure of the subgroups of Z/nZ has to do with the set of DIVISORS of n. Z/nZ is a cyclic group. the way an element [k] of Z/nZ behaves has everything to do with gcd(k,n). it's easy to see that if k divides n, then:

[k] has order n/k in Z/nZ.

what is perhaps more difficult to see, is that if gcd(k,n) = d, that [k] acts pretty much the same as [d] (insofar as the ADDITIVE structure of Z/nZ is concerned).

In this we are assuming that h is an additive group homomorphism and need to prove that h is also a multiplicative homomorphism and hence a ring homomorphism

so to interpret h(km) = h(km*1) = km*h(1) in the above, we have that the term km is k and m multiplied (ring multiplication operation) but the operation * means km additions of 1 ie 1+1 +1 + .... km times and hence your notational differentiation of the operations writing km not k*m for the ring multiplication of k and m and km*1 for km additions of 1

and hence what is happening when we write h(km*1) = km*h(1) we are smply using the relation or equation from the additive group homomorphism h(k) = h(k*1) = [k*h(1)